Are local diffeomorphisms Fredholm maps with index zero? [closed]

Question

Does this statement correct? if it does how we can prove it. In Banach spaces a map is local diffeomorphism if and only if it is a Fredholm map of index zero with no critical points?

Answer 1

If a map is a local diffeomorphism, then it is differentiable and has a differentiable inverse, so by the chain rule the derivative is a linear isomorphism of topological vector spaces, and therefore has 0 kernel and 0 cokernel, and is therefore Fredholm with index zero. This answers the question in the title. On the other hand, if the differential at a point is Fredholm, it has finite dimensional kernel and cokernel, but those might both be positive in dimension, and any such linear map is a differentiable map. So for example a 2x2 matrix with 1 dimensional kernel and 1 dimensional cokernel has Fredholm index zero but is not a local diffeomorphism. If in addition there are no critical points, i.e. the differential is everywhere surjective, then the cokernel is zero, so if the index is zero, then the kernel is also zero, and so the differential is a linear isomorphism. If the differential is Fredholm, it is also bounded, with bounded inverse modulo compact operators, but therefore bounded inverse (by the bounded inverse theorem), so an isomorphism of topological vector spaces. We can then apply the inverse function theorem as proven by Lang.